已知数列{an}是各项均为为0的等差数列，Sn为其前n项和，且满足S2n-1=
（I）求an； （Ⅱ）数列{bn}满足bn=
|
（I）设数列{an}的首项为a1，公差为d，
在S2n-1=
an2中，令n=1，2，得1 2
即a12=2S1 a22=2S3
…3分a12=2a1 (a1+d)2=2(3a1+3d)
解得a1=2，d=4，d=-2（舍去），
∴an=4n-2…5分
（Ⅱ）由（I）得bn=
…7分2n-1，n为奇数 2n-3，n为偶数
∴T2n=1+2×2-3+22+2×4-3+24+…+22n-2+2×2n-3…9分
=1+22+24+…+22n-2+4（1+2+…+n）-3n
=
+4•1-4n 1-4
-3nn(n+1) 2
=
-4n 3
+2n2-n…12分1 3
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